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Wolfram Language User Interface

Define a function with subscript expression?

Alex Graham

Posted 5 years ago

Hello, Tried this code but it dosent work. f[p0_, p1_] := Module[{Subscript[p, 0] = p0, Subscript[p, 1] = p1}, 1/(2 bb n (Subscript[p, 0] + (-1 + n) Subscript[p, 1])) \!\(\SubsuperscriptBox[\(p\), \(0\), \(-n\)]\) (-2 (-100 + d2) \!\(\SubsuperscriptBox[\(p\), \(0\), \(2\)]\) + 2 bb (d2 + 100 (-1 + n)) \!\(\SubsuperscriptBox[\(p\), \(0\), \(2 + n\)]\) - bb (-1 + n) n \!\(\SubsuperscriptBox[\(p\), \(0\), \(n\)]\) Subscript[p, 1] (200 + (-100 + 2 d1 - d2) Subscript[p, 1]) + 2 bb n \!\(\*SubsuperscriptBox[\(p\), \(0\), \(1 + n\)]\) (-100 + (-d1 + d2 + 100 (-1 + n)) Subscript[p, 1]))];

Hello, Tried this code but it dosent work.

f[p0_, p1_] := 
  Module[{Subscript[p, 0] = p0, Subscript[p, 1] = p1}, 
   1/(2 bb n (Subscript[p, 0] + (-1 + n) Subscript[p, 1])) 
\!\(\*SubsuperscriptBox[\(p\), \(0\), \(-n\)]\) (-2 (-100 + d2) 
\!\(\*SubsuperscriptBox[\(p\), \(0\), \(2\)]\) + 
      2 bb (d2 + 100 (-1 + n)) 
\!\(\*SubsuperscriptBox[\(p\), \(0\), \(2 + n\)]\) - bb (-1 + n) n 
\!\(\*SubsuperscriptBox[\(p\), \(0\), \(n\)]\) Subscript[p, 
       1] (200 + (-100 + 2 d1 - d2) Subscript[p, 1]) + 2 bb n 
\!\(\*SubsuperscriptBox[\(p\), \(0\), \(1 + 
         n\)]\) (-100 + (-d1 + d2 + 100 (-1 + n)) Subscript[p, 1]))];

POSTED BY: Alex Graham

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Hans Dolhaine

Hans Dolhaine, retired

Posted 5 years ago

What do you mean? Perhaps this? f[p0_, p1_] := Module[{}, Subscript[p, 0] = p0; Subscript[p, 1] = p1; 1/(2 bb n (Subscript[p, 0] + (-1 + n) Subscript[p, 1])) \! \SubsuperscriptBox[\(p\), \(0\), \(-n\)] (-2 (-100 + d2) \! \SubsuperscriptBox[\(p\), \(0\), \(2\)] + 2 bb (d2 + 100 (-1 + n)) \! \SubsuperscriptBox[\(p\), \(0\), \(2 + n\)] - bb (-1 + n) n \! \SubsuperscriptBox[\(p\), \(0\), \(n\)] Subscript[p, 1] (200 + (-100 + 2 d1 - d2) Subscript[p, 1]) + 2 bb n \! \*SubsuperscriptBox[\(p\), \(0\), \(1 + n\)] (-100 + (-d1 + d2 + 100 (-1 + n)) Subscript[p, 1]))]; f[a, b] Out[2]= (1/(2 bb (a + b (-1 + n)) n))a^-n (-2 a^2 (-100 + d2) + 2 a^(2 + n) bb (d2 + 100 (-1 + n)) + 2 a^(1 + n) bb (-100 + b (-d1 + d2 + 100 (-1 + n))) n - a^n b bb (200 + b (-100 + 2 d1 - d2)) (-1 + n) n)

What do you mean?

Perhaps this?

f[p0_, p1_] := 
    Module[{},
   Subscript[p, 0] = p0;
    Subscript[p, 1] = p1; 
      1/(2 bb n (Subscript[p, 0] + (-1 + n) Subscript[p, 1])) 
    \!
\*SubsuperscriptBox[\(p\), \(0\), \(-n\)] (-2 (-100 + d2) 
       \!
\*SubsuperscriptBox[\(p\), \(0\), \(2\)] + 
            2 bb (d2 + 100 (-1 + n)) 
       \!
\*SubsuperscriptBox[\(p\), \(0\), \(2 + n\)] - bb (-1 + n) n 
       \!
\*SubsuperscriptBox[\(p\), \(0\), \(n\)] Subscript[p, 
               1] (200 + (-100 + 2 d1 - d2) Subscript[p, 1]) + 2 bb n 
       \!
\*SubsuperscriptBox[\(p\), \(0\), \(1 + 
         n\)] (-100 + (-d1 + d2 + 100 (-1 + n)) Subscript[p, 1]))];


f[a, b]

Out[2]= (1/(2 bb (a + b (-1 + n)) n))a^-n (-2 a^2 (-100 + d2) + 
   2 a^(2 + n) bb (d2 + 100 (-1 + n)) + 
   2 a^(1 + n) bb (-100 + b (-d1 + d2 + 100 (-1 + n))) n - 
   a^n b bb (200 + b (-100 + 2 d1 - d2)) (-1 + n) n)

POSTED BY: Hans Dolhaine

Paul Abbott

Paul Abbott, Analytica International

Posted 5 years ago

Using built-in `Notation` package you can work directly with subscripted variables everywhere, including as arguments to functions.

POSTED BY: Paul Abbott

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